Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:

import java.io.BufferedReader;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class C {


	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();
		int arr[][] = new int[n][m];
		int arr2[][] = new int[n][m];
		
		for(int i=0;i<n;i++)
			for(int j=0;j<m;j++)
				arr2[i][j] = arr[i][j] = sc.nextInt();
		
		StringBuilder sb1 = new StringBuilder();
		int sum1 = 0;
		for(int i=0;i<n;i++)
		{
			int min = 1000;
			for(int j=0;j<m;j++)
				min = Math.min(min, arr[i][j]);
			if(min > 0)
			{
				for(int j=0;j<m;j++)
					arr[i][j]-=min;
				for(int k=0;k<min;k++)
					sb1.append("row ").append(i+1).append("\n");
				sum1+=min;
			}
		}
		for(int j=0;j<m;j++)
		{
			int min = 1000;
			for(int i=0;i<n;i++)
				min = Math.min(min, arr[i][j]);
			if(min > 0)
			{
				for(int i=0;i<n;i++)
					arr[i][j]-=min;
				for(int k=0;k<min;k++)
					sb1.append("col ").append(j+1).append("\n");
				sum1+=min;
			}
		}
		
		
		StringBuilder sb2 = new StringBuilder();
		int sum2 = 0;
		for(int j=0;j<m;j++)
		{
			int min = 1000;
			for(int i=0;i<n;i++)
				min = Math.min(min, arr2[i][j]);
			if(min > 0)
			{
				for(int i=0;i<n;i++)
					arr2[i][j]-=min;
				for(int k=0;k<min;k++)
					sb2.append("col ").append(j+1).append("\n");
				sum2+=min;
			}
		}
		for(int i=0;i<n;i++)
		{
			int min = 1000;
			for(int j=0;j<m;j++)
				min = Math.min(min, arr2[i][j]);
			if(min > 0)
			{
				for(int j=0;j<m;j++)
					arr2[i][j]-=min;
				for(int k=0;k<min;k++)
					sb2.append("row ").append(i+1).append("\n");
				sum2+=min;
			}
		}
		
		PrintWriter pw = new PrintWriter(System.out);
		if(sum1 < sum2)
		{
			for(int i=0;i<n;i++)
				for(int j=0;j<m;j++)
					if(arr[i][j] != 0)
					{
						System.out.println(-1);
						return;
					}
			pw.println(sum1);
			pw.print(sb1);
		}
		else
		{
			for(int i=0;i<n;i++)
				for(int j=0;j<m;j++)
					if(arr2[i][j] != 0)
					{
						System.out.println(-1);
						return;
					}
			pw.println(sum2);
			pw.print(sb2);
		}
		
		pw.flush();


	}

	static class Scanner{
		StringTokenizer st;BufferedReader br;
		public Scanner(InputStream s){	br = new BufferedReader(new InputStreamReader(s));}
		public Scanner(String s) throws FileNotFoundException{	br = new BufferedReader(new FileReader(new File(s)));}
		public String next() throws IOException 
		{while (st == null || !st.hasMoreTokens()) st = new StringTokenizer(br.readLine());return st.nextToken();}
		public int nextInt() throws IOException {return Integer.parseInt(next());}
		public long nextLong() throws IOException {return Long.parseLong(next());}
		public String nextLine() throws IOException {return br.readLine();}
		public double nextDouble() throws IOException {return Double.parseDouble(next());}
		public boolean ready() throws IOException {return br.ready();}
	}

}